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let-I-n-0-1-x-n-3-x-dx-1-calculate-lim-n-I-n-2-calculate-lim-n-n-I-n-

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Question Number 36413 by abdo.msup.com last updated on 01/Jun/18
let I_n = ∫_0 ^1 x^n (√(3+x))dx 1)calculate lim_(n→+∞) I_n 2) calculate lim_(n→+∞) n I_n
letIn=∫01xn3+xdx1)calculatelimn→+∞In2)calculatelimn→+∞nIn
Commented by abdo.msup.com last updated on 04/Jun/18
1) we have 0≤x≤1 ⇒ 3≤3+x≤4 ⇒ (√3) ≤(√(3+x))≤2 ⇒ ∫_0 ^1 (√3) x^n dx≤I_n ≤∫_0 ^1 2x^n dx ⇒ ((√3)/(n+1)) ≤ I_n ≤ (2/(n+1)) →_(n→+∞) 0 so lim_(n→+∞) I_n =0 2) by parts I_n =[ (1/(n+1))x^(n+1) (√(3+x))]_0 ^1 −∫_0 ^1 (x^(n+1) /(n+1)) (1/(2(√(3+x))))dx =(2/(n+1)) −(1/(2(n+1))) ∫_0 ^1 (x^(n+1) /( (√(3+x))))dx⇒ nI_n = ((2n)/(n+1)) −(n/(2n+2)) ∫_0 ^1 (x^(n+1) /( (√(3+x))))dx but (√3)≤(√(3+x))≤2 ⇒ (1/2) ≤ (1/( (√(3+x))))≤(√3) ⇒ (x^(n+1) /2) ≤ (x^(n+1) /( (√(3+x)))) ≤ (√3) x^(n+1) ⇒ ∫_0 ^1 (x^(n+1) /2)dx≤ ∫_0 ^1 (x^(n+1) /( (√(3+x))))dx≤ (√3)∫_0 ^1 x^(n+1) dx ⇒ (1/(2(n+2))) ≤ ∫_0 ^1 (x^(n+1) /( (√(3+x))))dx≤ ((√3)/(n+1)) so lim_(n→+∞) ∫_0 ^1 (x^(n+1) /( (√(3+x))))dx =0 ⇒ lim n I_n =2 (n→+∞)
1)wehave0⩽x⩽1⇒3⩽3+x⩽4⇒3⩽3+x⩽2⇒∫013xndx⩽In⩽∫012xndx⇒3n+1⩽In⩽2n+1→n→+∞0solimn→+∞In=02)bypartsIn=[1n+1xn+13+x]01−∫01xn+1n+1123+xdx=2n+1−12(n+1)∫01xn+13+xdx⇒nIn=2nn+1−n2n+2∫01xn+13+xdxbut3⩽3+x⩽2⇒12⩽13+x⩽3⇒xn+12⩽xn+13+x⩽3xn+1⇒∫01xn+12dx⩽∫01xn+13+xdx⩽3∫01xn+1dx⇒12(n+2)⩽∫01xn+13+xdx⩽3n+1solimn→+∞∫01xn+13+xdx=0⇒limnIn=2(n→+∞)

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